2 The following data were obtained from a population of tiger snails: phenotype
ID: 191748 • Letter: 2
Question
2 The following data were obtained from a population of tiger snails: phenotype genotype Light Medium Dark Total number of individuals 1469 138 1612 Calculate the observed allele frequencies for this trait using the above data. a. b. Assuming the population is in Hardy-Weinberg equilibrium, calculate the expected genotype frequencies using the observed allele frequencies. Are the observed genotype frequencies significantly different the expected genotype frequencies? What can be concluded from these results? c.Explanation / Answer
Answer:
2a).
AA = 1469 = 2938 “A” alleles
Aa = 138 = 138 “A” alleles & 138 “a” alleles
aa = 5 = 10 “a” alleles
Total “A” alleles = 2938+138 = 3076
Total “a” alleles = 138+10 = 148
Total alleles = 2938 + 276 + 10 = 3224
The frequency of allele “A” = 3076/3224 = 0.95
The frequency of allele “a” = 148/3224 = 0.05
b).
The frequency of aa = q^2 = 0.05*0.05 = 0.0025
The frequency of Aa = 2pq = 2 * 0.95 * 0.05 = 0.095
The frequency of AA = p^2 = 0.95 * 0.95 = 0.9025
aa= 0.0025 * 1612 = 4
Aa = 0.095 * 1612 = 153
AA = 0.9025 * 1612 = 1455
c). Yes the observed genotype frequencies are significantly different the expected genotype frequencies.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.